#11 Measure your calibre

First note that $15^{23} + 23^{23} = (19 - 4)^{23} + (19 + 4)^{23}$ .

Now when we expand these binomials, a factor of $19$ will appear in every term of the expansions except the last of each one, so we end up with

$15^{23} + 23^{23} \equiv (-4)^{23} + 4^{23} \pmod{19} = \boxed{0}$ .

If a,b are positive integers then, a^23 + b^23 is always divisible by (a+b). So it is divisible by 19.